\(\dfrac{4}{7}-\dfrac{1}{7}.\dfrac{10}{13}\)
\(=\dfrac{4}{7}-\dfrac{10}{91}\)
\(=\dfrac{52}{91}-\dfrac{10}{91}\)
\(=\dfrac{42}{91}=\dfrac{6}{13}\)
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\(\dfrac{1}{8}+\dfrac{1}{6}:\dfrac{2}{3}\)
\(=\dfrac{1}{8}+\dfrac{1}{6}.\dfrac{3}{2}\)
\(=\dfrac{1}{8}+\dfrac{1}{4}\)
\(=\dfrac{1}{8}+\dfrac{2}{8}\)
\(=\dfrac{3}{8}\)
___________
\(\left(\dfrac{7}{9}-\dfrac{2}{18}\right):\dfrac{4}{7}\)
\(=\left(\dfrac{14}{18}-\dfrac{2}{18}\right).\dfrac{7}{4}\)
\(=\dfrac{12}{18}.\dfrac{7}{4}\)
\(=\dfrac{7}{6}\)
\(#WendyDang\)
Lời giải:
$\frac{4}{7}-\frac{1}{7}\times \frac{10}{13}=\frac{1}{7}\times 4-\frac{1}{7}\times \frac{10}{13}$
$=\frac{1}{7}\times (4-\frac{10}{13})=\frac{1}{7}\times \frac{42}{13}$
$=\frac{42}{7\times 13}=\frac{6\times 7}{7\times 13}=\frac{6}{13}$
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$\frac{1}{8}+\frac{1}{6}: \frac{2}{3}=\frac{1}{8}+\frac{1}{6}\times \frac{3}{2}=\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}$
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$(\frac{7}{9}-\frac{2}{18}): \frac{4}{7}=(\frac{7}{9}-\frac{1}{9})\times \frac{7}{4}$
$=\frac{6}{9}\times \frac{7}{4}=\frac{2}{3}\times \frac{7}{4}=\frac{7}{6}$
